Prove 1 + 2 + 3 ... + n = n(n+1)/2 - Mathematical Induction - Theory

part 2 - Addition - Theory - Serial order wise - Mathematical Induction
part 3 - Addition - Theory - Serial order wise - Mathematical Induction

 

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Prove 1 + 2 + 3 + ……. + n = (š§(š§+šŸ))/šŸ for n, n is a natural number Step 1: Let P(n) : (the given statement) Let P(n): 1 + 2 + 3 + ……. + n = (n(n + 1))/2 Step 2: Prove for n = 1 For n = 1, L.H.S = 1 R.H.S = (š‘›(š‘› + 1))/2 = (1(1 + 1))/2 = (1 Ɨ 2)/2 = 1 Since, L.H.S. = R.H.S ∓ P(n) is true for n = 1 Step 3: Assume P(k) to be true and then prove P(k + 1) is true Assume that P(k) is true, P(k): 1 + 2 + 3 + ……. + k = (š‘˜(š‘˜ + 1))/2 We will prove that P(k + 1) is true. P(k + 1): 1 + 2 + 3 +……. + (k + 1) = ((k + 1)( (k + 1) + 1))/2 P(k + 1): 1 + 2 + 3 +…….+ k + (k + 1) = ((š¤ + šŸ)(š¤ + šŸ))/šŸ We have to prove P(k + 1) is true Solving LHS 1 + 2 + 3 +…….+ k + (k + 1) From (1): 1 + 2 + 3 + ……. + k = (š‘˜(š‘˜ + 1))/2 = (š’Œ(š’Œ + šŸ))/šŸ + (k + 1) = (š‘˜(š‘˜ + 1) + 2(š‘˜ + 1))/2 = ((š’Œ + šŸ)(š’Œ + šŸ))/šŸ = RHS ∓ P(k + 1) is true when P(k) is true Step 4: Write the following line Thus, By the principle of mathematical induction, P(n) is true for n, where n is a natural number

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